Hello,

I'm in minor confusion over slopes involving perpendicular lines.

Consider a rectangular co-ordinate system. The scales I'm using are such that ten small squares on my graph paper represent 1 unit on the x-axis and 2 units on the y-axis. I've drawn a straight line $\displaystyle y = 2x-2$ which passes through $\displaystyle x=1 $ when $\displaystyle y=0$. I've now drawn another straight line where both lines intersect at x=1 and are perpendicular (corner-to-corner for each square on the graph paper for both straight lines).

Now I define the equation for the second straight line choosing arbitrary points:

$\displaystyle y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x{1}} \cdot (x - x_{1}) $

$\displaystyle y - 4 = \frac{8 - 4}{(-3) - (-1)} \cdot [x - (-1)] \Rightarrow y = -2x + 2 $

For the point of intersection we solve simultaneously, from which yields x=1. The solution for both equations holds.

Further, two straight lines are perpendicular - if and only if the product of their slopes equates to $\displaystyle (-1) $

Therefore, $\displaystyle m_{1} \cdot m_{2} = (-1) $

$\displaystyle (2)\cdot(-2) \neq (-1) $

Delving deeper, one now remembers such that if you 'somehow' embrace the former line and rotate it through an angle of $\displaystyle 90^{\circ}$ then you have gone up x units and have subsequently moved towards the left by way of y units across.

We now define the equation for this perpendicular line from which is $\displaystyle y = - \frac{1}{2}x + 1 $

Therefore, $\displaystyle m_{1} \cdot m_{2} = (-1) $

$\displaystyle (2)\cdot(-\frac{1}{2}) = (-1) $

I now explain my confusion. This is an absolute load of horse radish. A perpendicular line can only have a negative reciprocal for a slope if one inverts the axes. In other words, the x-axis becomes the y-axis and on the converse the y-axis becomes the x-axis.

What this tells me is that these lines cannot be constructed in an identical rectangular co-ordinate system in order to have the product of their slopes equate to (-1).

For example, if I plot a line by means of subbin' in x values for $\displaystyle y = -\frac{1}{2}x + 1 $ using the former co-ordinate system I used to construct $\displaystyle y = 2x - 2$ they're not perpendicular.

What am I missing?

Thank you for your attention.